What is (\partial)/(\partial x)(sin(xy)z^2) ?

The answer to (\partial)/(\partial x)(sin(xy)z^2) is z^2cos(xy)y

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(\partial)/(\partial x)(sin(xy)z^2)

\frac{\partial}{\partial x} (sin (x y) z^{2})

z^{2} cos (x y) y

Solution steps

\frac{\partial}{\partial x} (sin (x y) z^{2})

Treat

y, z

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= z^{2} \frac{\partial}{\partial x} (sin (x y))

Apply the chain rule:

cos (x y) \frac{\partial}{\partial x} (x y)

= cos (x y) \frac{\partial}{\partial x} (x y)

\frac{\partial}{\partial x} (x y) = y

= z^{2} cos (x y) y

y = \frac{- 4 x}{x ^{2} + 1}, (0, 0)

x y^{^{'}} - 4 y = - 2 x

\frac{\partial}{\partial x} (x arccos (x))

\int cos (x) \cdot x d x

\int \frac{2 d t}{t ^{2}} d x

What is (\partial)/(\partial x)(sin(xy)z^2) ?
The answer to (\partial)/(\partial x)(sin(xy)z^2) is z^2cos(xy)y